We only have to complete the proof of the
integrality statement: \(c\in R, H(x)\in R[x]\) 
(left to the reader in [1]}). Suppose that \(c\notin R\) . Then
\(c^{-1}\in \pi R\) . As
\(\frac{1}{x}-c^{-1}=-(c^{-1}u)^2 x_1^{\prime }, \quad \text{where }x_1^{\prime }=\frac{x_1}{(c^{-1}u^2)x_1+1},\) 
by [2]}, Lemma 5.1, we have \(v(\Delta )< v(\Delta _1)\)  (with the notation
of op. cit., \(d=2v(c^{-1}u)>0\)  and the point \(p\)  is the pole of \(x\) 
in \(W_k\)  which is a smooth point).
It remains to
prove that \(H(x)\in R[x]\)  and \(\deg H\le g\) . Without loss of
generality we can suppose that \(c=0\) . We then have
\( P(x)+Q(x)H(x)+H(x)^2=u^{2(2g+1)}P_1(x_1)=u^{2(2g+1)}P_1(x/u^2)\in R[x].\) 
This implies that \(H(x)\in R[x]\)  using the Gauss valuation on \(K(x)\) 
associated to \(v_\mathfrak {p}\) . Finally, the same equality implies that \(\deg H\le g\) .
